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TCS latest pattern previous placement questions - 21

1. How many of the numbers x (x being integer) with 10<= x<= 99 are 18 more than the sum of their digitsa. 9b. 12c. 18d. 10
Answer: d
Explanation:
Let the number be ab. So given that
⇒10a + b = 18 + a + b
⇒9a = 18
⇒a = 2
So 20, 21, ... upto 29 there are total 10 numbers possible.

2. Apples cost L rupees per kilogram for the first 30 kilograms and Q per kilogram for each additional kilogram. If the price paid for 33 kilograms of Apples is Rs.1167 and for 36 kilograms of apples if Rs.1284, then the cost of the first 10 kgs of apples is:a. Rs.117b. Rs.350c. Rs.281d. Rs.1053
Answer: b
Explanation:
Given that
30L + 3Q = 1167
30L + 6Q = 1284
Solving we get Q = 39, L = 35
So cost of first 10 kgs of apples = 35 × 10 = 350

5. University of Vikramsila has enrolled nine PhD candidates. Babu, Chitra, Dheeraj , Eesha, Farooq,Gowri , Hameed, Iqbal, Jacob.-Farooq and Iqbal were enrolled on the same day as each other, and no one else was enrolled that day.-Chitra and gowri were enrolled on the same day as each other, and no one else was enrolled that day.-On each of the other days of hiring , exactly one candidate was enrolled.-Eesha was enrolled before Babu.-Hameed was enrolled before Dheeraj-Dheeraj was enrolled after Iqbal but before Eesha-Gowri was enrolled after both Jacob and Babu-Babu was enrolled before JacobWho were the last two candidates to be enrolled?a. Babu and Gowrib. Eesha and Jacobc. Babu and Chitrad. Gowri and Chitra
Answer: d
Explanation:
Given that
1. Easha < Babu
2. Hameed < Dheeraj
3. Iqbal < Dheeraj < Easha
4. Jacob/Babu < Gowri
5. Babu < Jacob
from 1 and 5, Easha was before Babu and Jacob so she cannot be in the last two. Option B ruled out
from 4 and 5, babu is before Jacob and Gowri so he cannot be in the last two. Options a, c ruled out.
So option d is correct.

6. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spade. Find the probability of the lost card being a spade.a. 10/50b. 10/53c. 11/50d. 11/53
Answer:
Explanation:

Let $S$ and ${S^1}$ be the respective events of choosing a spade and a card which is not spade. Let A denote drawing two spades. Out of 52 cards, 13 are spade and 39 cards are not spade.
P($S$) = 13/52 = 1/4
P(${S^1}$) = 39/52 = 3/4
We first calculate the total probability of drawing two spades when the missing card is a spade and the missing card is not a spade.
Total probability = $P(A)$ = $P(S \cap A) + P({S^1} \cap A)$ = $P(S).P\left( {\dfrac{A}{S}} \right) + P({S^1}).P\left( {\dfrac{A}{{{S^1}}}} \right)$
When one spade is lost, there are 12 spades out of 51 cards. Two cards can be drawn out of 12 spade cards in ${}^{12}{C_2}$ ways. Similarly, 2 cards can be drawn out of 51 cards in ${}^{51}{C_2}$ ways.
Probability of drawing 2 spades when one spade is lost = $\dfrac{{{}^{12}{C_2}}}{{{}^{51}{C_2}}}$ = $\dfrac{{22}}{{425}}$
$P(S \cap A) = P(S).P\left( {\dfrac{A}{S}} \right)$ = $\dfrac{1}{4} \times \dfrac{{22}}{{425}}$
When the lost card is not spade, there are 13 spades out of 51 cards. Two cards can be drawn out of 13 spades in ${{}^{13}{C_2}}$ ways whereas 2 cards can be drawn out of 51 cards in ${{}^{51}{C_2}}$ ways.
The probability of getting two cards, when one card is lost which is not spade, is given by $P\left( {\dfrac{A}{{{S^1}}}} \right)$
$P\left( {\dfrac{A}{{{S^1}}}} \right)$ = $\dfrac{{{}^{13}{C_2}}}{{{}^{51}{C_2}}}$ = $\dfrac{{26}}{{425}}$
$P({S^1} \cap A) = P({S^1}).P\left( {\dfrac{A}{{{S^1}}}} \right)$ = $\dfrac{3}{4} \times \dfrac{{26}}{{425}}$
The probability that the lost card is spade given that two spades are drawn = $P\left( {\dfrac{S}{A}} \right)$ = $\dfrac{{P(S \cap A)}}{{P(A)}}$ = $\dfrac{{P(S).P(A/S)}}{{P(S).P(A/S) + P({S^1}).P(A/{S^1})}}$ = $\dfrac{{1/4 \times 22/425}}{{1/4 \times 22/425 + 3/4 \times 26/425}}$ = 11/50

7. There are two bags containing white and black balls. In the first bag there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black.a. 21/154b. 7/54c. 21/77d. 41/77
Answer:
Explanation:
Probability = $\dfrac{1}{2} \times \dfrac{{{}^6{C_1}}}{{{}^{14}{C_1}}} + \dfrac{1}{2} \times \dfrac{{{}^7{C_1}}}{{{}^{11}{C_1}}}$ = $\dfrac{{41}}{{77}}$

8. A bag contains 1100 tickets numbered 1, 2, 3, ... 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?a. 291/1100b. 292/1100c. 290/1100d. 301/1100
Answer: c
Explanation:
Numbers which dont have 2 from 1 to 9 = 8
Numbers which dont have 2 from 10 to 99:
Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with 9 ways. Total 72 numbres.
Numbers which dont have 2 from 100 to 999 =_ _ _ = 8 × 9 × 9 = 648
Numbers which dont have 2 from 1000 to 1099 =10_ _ = 9 × 9 = 81
Finally 1100 does not have 2. So 1.
Total number with no 2 in them = 8 + 72 + 648 + 81 + 1= 810
Tickets with 2 in them = 1100 - 810 = 290
Required probability = 290 / 1100

9. In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.a) 1565 b) 2256 c) 2456 d) 1243
Answer: b
Explanation:
Maximum 3 men can be played which means there can be 0, 1, 2, 3 men in the team.
${(^5}{C_0}{ \times ^{11}}{C_{11}}) + {(^5}{C_1}{ \times ^{11}}{C_{10}})$ + ${(^5}{C_2}{ \times ^{11}}{C_9}) + {(^5}{C_3}{ \times ^{11}}{C_8})$ = 2256